Comparing complex eigenvectors if they are the same.

Question

You have two normalized eigenvectors $V_1$ and $V_2$ in $C^n$. Even if $V_1 - V_2 \neq 0$, the two vectors may still represent the same eigenvector, since $a V$ for any complex number $a$ with unit length, is the same eigenvector as $V$. One way to compare them is to normalize the two vectors in a new way: Divide each by its first non-zero component. Hence each vector will begin $[1,\ldots]$, unless the first component was zero, in which case it would start with $[0,1,\ldots]$, etc. Then, if $V_1 - V_2 = 0$, one can conclude that the two vectors represent the same eigenvector. Is there another way to compare them?

Answer 1

You'll have $|\langle v_1,v_2 \rangle | = 1$ if and only if they are parallel (i.e. $v_1 = a v_2$, $a \in \Bbb C$, $|a| = 1$). One direction is clear.

On the other hand, suppose $v_1 \not\parallel v_2$. Then $v_1 = c_1 v_2 + c_2 v_2^\perp$ for some unit length vector $v_2^\perp$ that is orthogonal to $v_2$, and such that $|c_1|^2 + |c_2|^2 = 1$. Thus $|c_1| < 1$ if $c_2 \neq 0$.

Then

$$|\langle v_1,v_2 \rangle| = |\langle c_1 v_2, v_2 \rangle| = |c_1| < 1,$$

proving the other direction.

This is basically the second half of the statement of Cauchy-Schwarz:

$$ |\langle v,w\rangle| \leq \|v\|\|w\| $$

with equality if and only if $v \parallel w$.

Answer 2

You just need to find a scalar $c$ such that $v_1 = c v_2$.

Let $k$ be the first index for which $[v_2]_k \neq 0$ and let $c = { [v_1]_k \over [v_2]_k }$. If $|c| =1$ (or close) then check $\|v_1-c v_2\|$.